\newproblem{lay:1_3_31}{
   % Problem identification
	 \begin{large}
	  \hspace{\fill}\newline
     \textbf{Lay, 1.3.31}
	 \end{large}
	 \\
   \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

   % Problem statement
   A thin triangular, metal plate of uniform density and thickness has vertices $\mathbf{v}_1=(0,1)$,
	 $\mathbf{v}_2=(8,1)$, and $\mathbf{v}_3=(2,4)$
	 \begin{center}
		\includegraphics[scale=0.4]{Tema1/lay_1_3_31.eps}
	 \end{center}
	 and the mass of the plate is 3g.
	\begin{enumerate}[a]
		\item Find the $(x,y)$-coordinates of the center of mass of the plate. This ``balanced point'' of the plate 
		      coincides with the center of mass of a system consisting of three 1-gram point masses located at the
					vertices of the plate.
		\item Determine how to distribute an additional mass of 6g at the three vertices to move the balance point of
		      the plate to (2,2).
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a]
	\item Let us calculate the total mass
		 \begin{center}
			 $m=m_1+m_2+m_3=1+1+1=3 g$
		 \end{center}
		Now, the center of mass
		 \begin{center}
			 $\overline{\mathbf{v}}=\frac{1}{m}(m_1\mathbf{v}_1+m_2\mathbf{v}_2+m_3\mathbf{v}_3)=\frac{1}{3}\left(
				 1\begin{pmatrix}0\\1\end{pmatrix}+1\begin{pmatrix}8\\1\end{pmatrix}+1\begin{pmatrix}2\\4\end{pmatrix}\right)=
				\begin{pmatrix}\frac{10}{3}\\2\end{pmatrix}$
		 \end{center}
		\item If we now want to shift the center of masses, let us define as $w_1$, $w_2$ and $w_3$ the masses to be added to each one of the vertices, with the constraint
		 \begin{center}
			 $w_1+w_2+w_3=6$
		 \end{center}
		The new center of masses will be
		 \begin{center}
			 $\overline{\mathbf{v}}=\frac{1}{m+6}((m_1+w_1)\mathbf{v}_1+(m_2+w_2)\mathbf{v}_2+(m_3+w_3)\mathbf{v}_3)=\frac{1}{9}\left(
				 \begin{pmatrix}0\\1+w_1\end{pmatrix}+\begin{pmatrix}8(1+w_2)\\1+w_2\end{pmatrix}+\begin{pmatrix}2(1+w_3)\\4(1+w_3)\end{pmatrix}\right)=
				\begin{pmatrix}2\\2\end{pmatrix}$\\
				$\begin{pmatrix}\frac{10+8w_2+2w_3}{9}\\\frac{6+w_1+w_2+4w_3}{9}\end{pmatrix}=\begin{pmatrix}2\\2\end{pmatrix}$
		 \end{center}
		which gives us the equation system
		 \begin{center}
			 $\begin{array}{rcl}
			   w_1+w_2+w_3&=&6\\
			   8w_2+2w_3&=&8\\
			   w_1+w_2+4w_3&=&12\\
				\end{array}$
		 \end{center}
		whose solution is $w_1=3.5$g, $w_2=0.5$g, $w_3=2$g.
	 \end{enumerate}
}

\useproblem{lay:1_3_31}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
